Find $f^{-100}g^{146}f^{301}$ where
$$f = \begin{pmatrix} 1 & 2 & 3& 4 & 5 & 6 & 7 \\ 3 & 1 & 5 & 7 & 2 & 6 & 4\end{pmatrix}, \\ g = \begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ 3 & 1 & 7 & 6 & 4 & 5 & 2\end{pmatrix}. $$ can someone please help with a step by step guide for this as I can't find any examples anywhere and Im really confused. I can put it in cycle notation and find order but don't know how to use that information to find answer. Thanks
Edit- cycle notation for f = (1352)(47) and order is lcm(4,2)=4 cycle notation for g = (1372)(465) and order is lcm(4,3)=12
Hint:
To calculate easily powers of a permutation, decompose it first as a product of disjoint cycles.
E.g., $\;f=(1\,3\,5\,2)(4\,7)$.
Now a cycle has order its length, and a product of disjoint cycles has order the l.c.m. of the orders of their orders, i.e. the l.c.m. of thelengths of each cycle. Thus, $f$ has order $\operatorname{lcm}(4,2)=4$.
On the other hand, $f^2$ is easily obtained by the rule than an element in the cycle maps onto the second next, as if these elements were distributed clockwise on a circle: so that $\;f^2=(1\,5)(2\,3)$. Similarly, for $f^3$, an element maps onto the third next: $\;f^3=(1\,2\,5\,3)$. You can observe it is the same as $f^{-1}$, which is obtained mapping each element onto the next o,e counterclockwise.
Last hint: as $f^4=1$ $ ;\ f^{301} =f^{301\bmod 4} =f^1$.
Can you perform all the calculations with these elements?